Ask question

Ask question

All categories

3 years ago

11

A 7.5 kg block is placed on a table. if it's bottom surface area is 0.6m2, how much pressure does the block exert on the tableto

p?
A) 367.5 Pa
B) 122.5 Pa
C) 226.5 Pa
D) 73.5 Pa

1 answer:

xeze [42]3 years ago

3 0

The answer to this is B.

You might be interested in

A projectile is launched from the ground with an initial velocity of 12ms at an angle of 30° above the horizontal. The projectil

netineya [11]

$vi^{2}sin2thita/g =12^{2}sin2[30]/9.8=12.7$ Answer:

Explanation:

range is given as

6 0

3 years ago

Read 2 more answers

Three science questions help please!

Nimfa-mama [501]

6. D
7. D
8. B
let me know if you need clarification

4 0

4 years ago

The distance between adjacent nodes in a standing wave pattern in a length of string is 25.0 cm:A. What is the wavelength of wav

mina [271]

A) 50 cm

B) 10000 cm/s

Explanation

Step 1

A)

If you know the distance between nodes and antinodes then use this equation:

$\begin{gathered} \frac{\lambda}{2}=D \\ \text{where}\lambda\text{ is the wavelength} \\ D\text{ is the distance betw}een\text{ nodes} \end{gathered}$

then, let

$D=\text{ 25 cm }$

now, replace to find the wavelength

$\begin{gathered} \frac{\lambda}{2}=25 \\ \text{Multiply both sides by 2} \\ \frac{\lambda}{2}\cdot2=25\cdot2 \\ \lambda=50\text{ Cm} \end{gathered}$

so, the wavelength is

A) 50 cm

Step 2

The speed of a wave can be found using the equation

$v=\lambda f$

or velocity = wavelength x frequency,

then,let

$\begin{gathered} \lambda=50\text{ cm} \\ f=200\text{ Hz} \end{gathered}$

replace and evaluate

$\begin{gathered} v=\lambda f \\ v=50\text{ cm }\cdot200\text{ HZ} \\ v=10000\text{ }\frac{\text{cm}}{s} \end{gathered}$

so

B) 10000 cm/s

I hope this helps you

6 0

1 year ago

g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t

jok3333 [9.3K]

Answer:

(a) f = 0.58Hz

(b) vmax = 0.364m/s

(c) amax = 1.32m/s^2

(d) E = 0.1J

(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

$f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz$

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

$v_{max}=\omega A=2\pi f A$ (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

$v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}$

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

$a_{max}=\omega^2A=(2\pi f)^2 A$

$a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}$

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

$E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J$

The total energy is 0.1J

(e) The displacement as a function of time is:

$x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)$

6 0

3 years ago

suppose that 273 g of one of the substances listed above displaces 26 mL of water. What is the substance?

guajiro [1.7K]

<span>The unknown substance is silver. I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so 273 g / 26 mL = 10.5 g/mL Looking up a list of elements sorted by density, I see the following: 10.07 Actinium 10.22 Molybdenum 10.5 Silver 11.35 Lead And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.</span>

6 0

4 years ago

Read 2 more answers